To determine the density of the material of the beaker, we analyze the conditions under which the beaker floats or remains at the bottom.

### Given Data:
- Mass of the beaker, \( m = 390 \) g = \( 0.390 \) kg
- Interior volume of the beaker, \( V_{\text{int}} = 500 \) cm³ = \( 500 \times 10^{-6} \) m³
- The beaker floats if it is **less than half full**
- The beaker sinks if it is **more than half full**

### Step 1: Understand Floating and Sinking Conditions
A floating object experiences a buoyant force equal to its weight:

\[
F_{\text{buoyant}} = W_{\text{beaker}}
\]

The buoyant force is given by:

\[
F_{\text{buoyant}} = \rho_{\text{water}} V_{\text{displaced}} g
\]

where \( \rho_{\text{water}} \) is the density of water (\( 1000 \) kg/m³), and \( V_{\text{displaced}} \) is the volume of water displaced.

For the beaker to float when it is **less than half full**, it must be able to displace a volume of water equal to its total weight.

For the beaker to sink when it is **more than half full**, its total weight must exceed the buoyant force when the sink is filled to the beaker’s rim.

### Step 2: Define Key Variables
Let \( V_{\text{beaker}} \) be the volume of the material of the beaker itself. Since the **interior** volume is 500 cm³, the total volume of the beaker is:

\[
V_{\text{total}} = V_{\text{beaker}} + V_{\text{int}}
\]

Let \( \rho_{\text{beaker}} \) be the density of the material of the beaker.

### Step 3: Floating Condition
For the beaker to **just float when half full**, the total weight must equal the buoyant force:

\[
(m + m_{\text{water}}) g = \rho_{\text{water}} V_{\text{total}} g
\]

where \( m_{\text{water}} \) is the mass of the water inside the beaker when half full:

\[
m_{\text{water}} = \rho_{\text{water}} \times \frac{V_{\text{int}}}{2}
\]

Substituting values:

\[
m_{\text{water}} = 1000 \times \frac{500 \times 10^{-6}}{2}
\]

\[
m_{\text{water}} = 0.250 \text{ kg}
\]

Thus, the total mass when half full is:

\[
m_{\text{total}} = 0.390 + 0.250 = 0.640 \text{ kg}
\]

Since it just floats, the buoyant force must be:

\[
F_{\text{buoyant}} = 1000 \times V_{\text{total}} \times g
\]

\[
0.640 g = 1000 V_{\text{total}} g
\]

Canceling \( g \):

\[
0.640 = 1000 V_{\text{total}}
\]

\[
V_{\text{total}} = 0.00064 \text{ m}^3 = 640 \text{ cm}^3
\]

### Step 4: Solve for Density
Since \( V_{\text{total}} = V_{\text{beaker}} + V_{\text{int}} \), we solve for \( V_{\text{beaker}} \):

\[
V_{\text{beaker}} = 640 - 500 = 140 \text{ cm}^3 = 140 \times 10^{-6} \text{ m}^3
\]

The density of the beaker material is:

\[
\rho_{\text{beaker}} = \frac{m}{V_{\text{beaker}}}
\]

\[
\rho_{\text{beaker}} = \frac{0.390}{140 \times 10^{-6}}
\]

\[
\rho_{\text{beaker}} = 2785.7 \text{ kg/m}^3
\]

### Final Answer:
The density of the material of the beaker is approximately **2800 kg/m³**.