Work Done Moving Mass from Axis of Annular Disc to Infinity

To find the work required to move a unit mass from point $P$ on the axis of the annular disc to infinity, we first need to determine the gravitational potential at point $P$.

1. Surface Mass Density of the Annular Disc

The outer radius of the disc is $R_2 = 4R$ and the inner radius is $R_1 = 3R$.
The total surface area of the annular disc is:
$A = \pi R_2^2 - \pi R_1^2 = \pi (4R)^2 - \pi (3R)^2 = 7\pi R^2$

Thus, the surface mass density $\sigma$ of the disc is:
$\sigma = \frac{M}{A} = \frac{M}{7\pi R^2}$

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2. Gravitational Potential at Point $P$

Point $P$ lies on the central axis of the disc at a distance $z = 4R$ from its center.

Consider a thin circular ring element of radius $r$ and thickness $dr$ centered on the disc. The mass of this element is:
$dm = \sigma \cdot (2\pi r \, dr)$

The gravitational potential $dV$ at point $P$ due to this differential ring element is:
$dV = -\frac{G \, dm}{\sqrt{r^2 + z^2}} = -\frac{2\pi G \sigma r \, dr}{\sqrt{r^2 + z^2}}$

Integrating $dV$ over the entire annular disc from $r = 3R$ to $r = 4R$:
$V_P = -2\pi G \sigma \int_{3R}^{4R} \frac{r \, dr}{\sqrt{r^2 + z^2}}$

Using the integration rule $\int \frac{r \, dr}{\sqrt{r^2 + z^2}} = \sqrt{r^2 + z^2}$:
$V_P = -2\pi G \sigma \left[ \sqrt{r^2 + z^2} \right]_{3R}^{4R}$

Substitute $z = 4R$:
$V_P = -2\pi G \sigma \left( \sqrt{(4R)^2 + (4R)^2} - \sqrt{(3R)^2 + (4R)^2} \right)$
$V_P = -2\pi G \sigma \left( \sqrt{32 R^2} - \sqrt{25 R^2} \right) = -2\pi G \sigma (4\sqrt{2} R - 5R)$

Substituting the value of $\sigma = \frac{M}{7\pi R^2}$:
$V_P = -2\pi G \left( \frac{M}{7\pi R^2} \right) R (4\sqrt{2} - 5)$
$V_P = -\frac{2GM}{7R}(4\sqrt{2} - 5)$

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3. Calculation of Work Done

The work required by an external agent to take a unit mass ($m = 1$) from point $P$ to infinity ($r \to \infty$) slowly is given by the change in gravitational potential energy:
$W = U(\infty) - U(P) = m \left( V_\infty - V_P \right)$

Since $V_\infty = 0$ and $m = 1$:
$W = 0 - V_P = -V_P$

Substituting $V_P$:
$W = -\left( -\frac{2GM}{7R}(4\sqrt{2} - 5) \right) = \frac{2GM}{7R}(4\sqrt{2} - 5)$

Thus, the correct option is A.