Step 1: State the long fin heat loss result.
For an infinitely long fin the tip condition drops out and the heat carried away equals $Q = M \theta_b$ where $M = \sqrt{hPkA}$ comes from solving the fin equation $\dfrac{d^2\theta}{dx^2} = \dfrac{hP}{kA}\theta$. Since $h$, $k$, and the base temperature excess $\theta_b$ do not change on replacing the fin, only the shape factor $\sqrt{PA}$ decides the ratio.
Step 2: Try a concrete number to keep the algebra simple.
Let the square side and the circle diameter both equal $a = 1$ m (any value works since the ratio is dimensionless in $a$). Square: $P = 4(1) = 4$, $A = 1^2 = 1$, so $PA = 4$. Circle: $P = \pi(1) = \pi$, $A = \pi(1)^2/4 = \pi/4$, so $PA = \pi \times \pi/4 = \pi^2/4$.
Step 3: Form the ratio of the two heat rates.
$$\frac{Q_{before}}{Q_{after}} = \frac{\sqrt{(PA)_{square}}}{\sqrt{(PA)_{circle}}} = \sqrt{\frac{4}{\pi^2/4}} = \sqrt{\frac{16}{\pi^2}} = \frac{4}{\pi}$$
Step 4: Sanity check the number.
Since $\pi \approx 3.14$, the ratio is about $1.27$, so the square fin carries roughly 27 percent more heat than the circular one of the same characteristic size, which fits since a square packs more perimeter into the same area than a circle does.
Final Answer:
The before to after heat rate ratio comes out to $4/\pi$, the same result however you scale the side length.
\[ \boxed{\dfrac{4}{\pi}} \]