1. Set up the radius of the ice-coated ball

Let the thickness of ice at time $t$ be $x$ cm.

Since the iron ball has radius $10$ cm, the outer radius of the ice-coated sphere is $R = 10 + x.$

When the thickness of ice is $5$ cm, $R = 10+5=15\text{ cm}.$

2. Write the volume of ice

The volume of ice is the volume of the outer sphere minus the volume of the iron ball: $V = \frac{4}{3}\pi R^3 - \frac{4}{3}\pi (10)^3.$

Since $10^3$ is constant, differentiating with respect to time gives $\frac{dV}{dt} = 4\pi R^2 \frac{dR}{dt}.$

But $R=10+x$, so $\frac{dR}{dt} = \frac{dx}{dt}.$

Hence, $\frac{dV}{dt} = 4\pi R^2 \frac{dx}{dt}.$

3. Use the melting rate

The ice melts at a rate of $50\text{ cm}^3/\text{min}$, so the volume of ice is decreasing: $\frac{dV}{dt} = -50.$

At the instant when $x=5$, we have $R=15$. Substituting: $-50 = 4\pi (15)^2 \frac{dx}{dt}.$

So, $-50 = 4\pi \cdot 225 \frac{dx}{dt} = 900\pi \frac{dx}{dt}.$

Therefore, $\frac{dx}{dt} = \frac{-50}{900\pi} = -\frac{1}{18\pi}.$

4. Interpret the result

The negative sign shows that the thickness is decreasing. Therefore, the rate at which the thickness decreases is $\left|\frac{dx}{dt}\right| = \frac{1}{18\pi}\text{ cm/min}.$

5. Check options

The correct option is: $\boxed{\text{B } \frac{1}{18\pi}\text{ cm/min}}$

6. Comparison with stored answer

Stored correct answer: B

Our derived answer: B

So, the derived answer agrees with the stored answer.