A cylindrical vessel having diameter equal to its height is full of water which is poured into two identical cylindrical vessels with diameter 42 cm and height 21 cm, which are completely filled. Find the diameter of the cylindrical vessel.

Let the diameter of the original cylindrical vessel be \( D \) cm, and since the diameter equals the height, the height is also \( D \) cm. The volume of a cylinder is given by:

\[ V = \pi \times \text{radius}^2 \times \text{height} \]

For the original vessel:
- Radius \( r = \frac{D}{2} \)
- Height \( h = D \)

Volume of the original vessel:
\[ V_{\text{original}} = \pi \times \left(\frac{D}{2}\right)^2 \times D = \frac{\pi D^3}{4} \]

Now, for the two smaller vessels:
- Radius \( r_{\text{small}} = \frac{42}{2} = 21 \) cm
- Height \( h_{\text{small}} = 21 \) cm

Volume of one small vessel:
\[ V_{\text{small}} = \pi \times 21^2 \times 21 = \pi \times 441 \times 21 = 9261\pi \, \text{cm}^3 \]

The total volume of the two smaller vessels:
\[ V_{\text{total}} = 2 \times V_{\text{small}} = 2 \times 9261\pi = 18522\pi \, \text{cm}^3 \]

Since the original vessel's volume equals the total volume of the two smaller vessels:
\[ \frac{\pi D^3}{4} = 18522\pi \]

Simplify the equation:
\[ D^3 = 18522 \times 4 \]
\[ D^3 = 74088 \]

Take the cube root of both sides:
\[ D = \sqrt[3]{74088} \]

Approximating the cube root:
\[ D = 42 \, \text{cm} \]

Thus, the diameter of the original cylindrical vessel is **42 cm**.