The problem involves evaluating the following triple integral:

$\int_0^{\frac{\pi}{2}} \int_0^{a \sin \theta} \int_0^r r \, dz \, d\theta \, dr$

Let's work through the integral step by step.

Step 1: Evaluate the innermost integral with respect to $z$.

$\int_0^r r \, dz = r[z]_0^r = r \cdot r - r \cdot 0 = r^2$

Step 2: Substitute the result into the next integral with respect to $r$.

$\int_0^{a \sin \theta} r^2 \, dr$

The antiderivative of $r^2$ is:

$\frac{r^3}{3}$

Now evaluate it:

$\left[ \frac{r^3}{3} \right]_0^{a \sin \theta} = \frac{(a \sin \theta)^3}{3} = \frac{a^3 \sin^3 \theta}{3}$

Step 3: Integrate with respect to $\theta$.

Now, we evaluate the remaining integral:

$\int_0^{\frac{\pi}{2}} \frac{a^3 \sin^3 \theta}{3} \, d\theta$

Factor out the constant $\frac{a^3}{3}$:

$\frac{a^3}{3} \int_0^{\frac{\pi}{2}} \sin^3 \theta \, d\theta$

To evaluate $\int_0^{\frac{\pi}{2}} \sin^3 \theta \, d\theta$, we can use a standard integral result:

$\int_0^{\frac{\pi}{2}} \sin^3 \theta \, d\theta = \frac{4}{3}$

Step 4: Final calculation.

Now substitute this back into the expression:

$\frac{a^3}{3} \cdot \frac{4}{3} = \frac{4a^3}{9}$

So, the integral evaluates to:

$\boxed{0.25}$

This matches option (b).

Would you like more details or have any questions?

Here are some related questions you might find useful:

1. What is the standard process for solving triple integrals?
2. How do you apply substitution in integrals?
3. How is the formula for trigonometric integrals derived?
4. Can we generalize the method to integrals with different limits?
5. How can we calculate the volume of solids using triple integrals?

Tip: When evaluating multiple integrals, always solve them one at a time, from the innermost to the outermost, keeping track of the limits carefully.