Solution

The optical power of a thin lens $D$ and its focal length $F$ are calculated using the universal formula:
$$D = \frac{1}{F} = \left(\frac{n_{\text{l}}}{n_{\text{m}}} - 1\right) \left( \pm \frac{1}{R_1} \pm \frac{1}{R_2} \right)$$
where $n_{\text{l}}$ is the refractive index of the lens material, and $n_{\text{m}}$ is the refractive index of the surrounding medium.
The "$+$" sign in the brackets is used if the surface is convex, and the "$-$" sign if the surface is concave. The values of the radii of curvature $R_1$ and $R_2$ are substituted by their absolute values.
Since the medium is not specified in the problem, we assume the lens is in air ($n_{\text{m}} = 1$). The refractive index of glass is $n_{\text{l}} = 1.6$.

Part a)

The lens is biconvex, so both of its surfaces are convex. We place a "$+$" sign before both terms with radii. According to the problem, $R_1 = R_2 = R = 0.3 \text{ m}$.
Optical power of the lens:
$$D = \left(\frac{n_{\text{l}}}{1} - 1\right) \left( \frac{1}{R} + \frac{1}{R} \right) = \frac{2(n_{\text{l}} - 1)}{R}$$
Substituting the values:
$$D = \frac{2(1.6 - 1)}{0.3} = \frac{1.2}{0.3} = 4 \text{ D}$$
Focal length:
$$F = \frac{1}{D} = \frac{1}{4} = 0.25 \text{ m} = 25 \text{ cm}$$

Part b)

The lens is plano-spherical. Since the optical power of the lens is positive ($D = 1 \text{ D} > 0$), the lens is converging. It follows that its spherical surface must be convex (if it were concave, the lens would be diverging).
For a flat surface, the radius of curvature is $R_1 \to \infty$, so $\frac{1}{R_1} = 0$.
Taking into account the "$+$" sign for the second, convex surface, the lens formula takes the form:
$$D = (n_{\text{l}} - 1) \left( 0 + \frac{1}{R_2} \right) = \frac{n_{\text{l}} - 1}{R_2}$$
Expressing the required radius of curvature $R_2$:
$$R_2 = \frac{n_{\text{l}} - 1}{D}$$
Substituting the values:
$$R_2 = \frac{1.6 - 1}{1} = 0.6 \text{ m} = 60 \text{ cm}$$