Let's break down the problem systematically.

### Given data:
- Rotational inertia of the wheel: \( I = 0.750 \) kg·m²
- Mass of the wheel: \( m = 15.0 \) kg
- Radius of the wheel: \( R = 30.0 \) cm = 0.30 m
- Speed of the conveyor belt: \( v = 12.0 \) m/s
- Coefficient of kinetic friction: \( \mu_k = 0.600 \)

### Step 1: Understanding the Motion
Initially, the wheel is motionless and placed on the moving conveyor belt. Due to friction, it will begin to rotate while simultaneously undergoing translational motion. The goal is to determine how long it takes for the wheel to reach its final angular velocity, where it rolls without slipping.

When rolling without slipping, the final velocity of the center of mass \( v_f \) and final angular velocity \( \omega_f \) satisfy the condition:

\[
v_f = R \omega_f
\]

### Step 2: Force and Acceleration Analysis
The kinetic friction force \( f_k \) acts at the contact point between the tire and the conveyor belt:

\[
f_k = \mu_k m g
\]

Substituting the values:

\[
f_k = (0.600)(15.0)(9.81) = 88.29 \text{ N}
\]

This friction force causes both linear acceleration of the center of mass and angular acceleration about the wheel’s axis.

#### Linear Acceleration:
Newton's second law gives:

\[
m a = f_k
\]

\[
15.0 a = 88.29
\]

\[
a = \frac{88.29}{15.0} = 5.886 \text{ m/s}^2
\]

#### Angular Acceleration:
The torque due to friction about the wheel's center is:

\[
\tau = f_k R
\]

\[
\tau = (88.29)(0.30) = 26.49 \text{ N·m}
\]

Using Newton’s second law for rotation:

\[
I \alpha = \tau
\]

\[
(0.750) \alpha = 26.49
\]

\[
\alpha = \frac{26.49}{0.750} = 35.32 \text{ rad/s}^2
\]

### Step 3: Time to Achieve Rolling Without Slipping
Initially, the wheel's center of mass is at rest \( (v_0 = 0) \), and its angular velocity is also zero \( (\omega_0 = 0) \). The rolling condition is achieved when:

\[
v_f = R \omega_f
\]

From kinematics,

\[
v_f = a t
\]

\[
\omega_f = \alpha t
\]

Since \( v_f = R \omega_f \), substituting the expressions:

\[
(a t) = R (\alpha t)
\]

\[
t (a - R \alpha) = 0
\]

\[
t = \frac{v}{a + R\alpha}
\]

\[
t = \frac{12.0}{5.886 + (0.30)(35.32)}
\]

\[
t = \frac{12.0}{5.886 + 10.596} = \frac{12.0}{16.482}
\]

\[
t = 0.728 \text{ seconds}
\]

### Step 4: Length of Skid Mark
The length of the skid mark is the distance traveled by the conveyor belt in time \( t \), which is:

\[
d = v t
\]

\[
d = (12.0)(0.728)
\]

\[
d = 8.736 \text{ m}
\]

### Final Answer:
(a) The time required for the wheel to achieve its final angular velocity is **0.728 seconds**.
(b) The length of the skid mark on the conveyor surface is **8.736 meters**.