1. Magnetic field as a function of time

   The magnetic field is given by $B(t)=B_0 e^{-t/\tau}$ (the symbol in the exponent must be the time constant $\tau$).

2. Magnetic flux through the circular loop

   Since the loop is perpendicular to the magnetic field, the angle between area vector and $\vec B$ is $0^\circ$.

   Area of the loop: $A=\pi r^2$

   Hence magnetic flux is $\Phi(t)=BA=\pi r^2 B_0 e^{-t/\tau}$

3. Induced emf

   By Faraday’s law, $\mathcal E = -\frac{d\Phi}{dt}$

   So magnitude of emf is $\mathcal E(t)=\left|\frac{d}{dt}(\pi r^2 B_0 e^{-t/\tau})\right|$ $\mathcal E(t)=\pi r^2 B_0 \cdot \frac{1}{\tau} e^{-t/\tau}$

   Thus, $\mathcal E(t)=\frac{\pi r^2 B_0}{\tau}e^{-t/\tau}$

4. Current in the loop

   Using Ohm’s law, $I(t)=\frac{\mathcal E(t)}{R}=\frac{\pi r^2 B_0}{\tau R}e^{-t/\tau}$

5. Power dissipated as heat

   Instantaneous Joule heating power is $P(t)=I^2R=\frac{\mathcal E^2}{R}$

   Therefore, $P(t)=\frac{1}{R}\left(\frac{\pi r^2 B_0}{\tau}e^{-t/\tau}\right)^2$ $P(t)=\frac{\pi^2 r^4 B_0^2}{\tau^2 R}e^{-2t/\tau}$

6. Total heat generated for $t=0$ to $t\to\infty$

   $Q=\int_0^{\infty} P(t)\,dt$

   $Q=\frac{\pi^2 r^4 B_0^2}{\tau^2 R}\int_0^{\infty} e^{-2t/\tau}\,dt$

   Now, $\int_0^{\infty} e^{-2t/\tau}dt = \frac{\tau}{2}$

   Hence, $Q=\frac{\pi^2 r^4 B_0^2}{\tau^2 R}\cdot \frac{\tau}{2}$ $Q=\frac{\pi^2 r^4 B_0^2}{2\tau R}$

7. Matching with options

   The correct expression is $\boxed{\frac{\pi^2 r^4 B_0^2}{2\tau R}}$

   This matches Option B.

Note: In the typed options, the power of $r$ appears as $r^4$ in the numerator only if read properly from the scan/context. Physically and dimensionally also, the result must contain $r^4$ in the numerator, not denominator.