1. Set up the variables

Let:

• $r$ = radius of the chocolate ball
• $x$ = thickness of the ice-cream layer

Then the outer radius of the chocolate + ice-cream is $R = r + x$

The volume of ice-cream alone is $V = \frac{4}{3}\pi R^3 - \frac{4}{3}\pi r^3 = \frac{4}{3}\pi \big((r+x)^3-r^3\big)$

Since the chocolate ball does not melt, $r$ is constant.

2. Differentiate with respect to time

$\frac{dV}{dt} = 4\pi (r+x)^2 \frac{dx}{dt}$

3. Substitute the given data

When the thickness is $1$ cm: $x=1$

The ice-cream melts at the rate $81\,\text{cm}^3/\text{min}$, so $\frac{dV}{dt} = -81$

The thickness decreases at the rate $\dfrac{1}{4\pi}$ cm/min, so $\frac{dx}{dt} = -\frac{1}{4\pi}$

Substitute into $\frac{dV}{dt} = 4\pi (r+x)^2 \frac{dx}{dt}$

We get $-81 = 4\pi (r+1)^2\left(-\frac{1}{4\pi}\right)$

This simplifies to $-81 = -(r+1)^2$ $(r+1)^2 = 81$

Since radius is positive, $r+1=9 \implies r=8$

4. Find the surface area of the chocolate ball

Surface area of the chocolate ball is $4\pi r^2 = 4\pi(8^2)=4\pi(64)=256\pi$

5. Match with the options

Thus the required surface area is $\boxed{256\pi}$

So the correct option is D.