Given at 298 K :
$$E^\ominus_{\text{Fe}^{2+}/\text{Fe}} = X$$ Volt
$$E^\ominus_{\text{Fe}^{3+}/\text{Fe}} = Y$$ Volt
The $$E^\ominus_{\text{Fe}^{3+}/\text{Fe}^{2+}}$$ in Volt at 298 K is given by :

Standard electrode potentials cannot be added directly. First, convert them into standard Gibbs free energy changes using

$$\Delta G^\circ=-nFE^\circ$$

Given:

$$\mathrm{Fe^{2+}+2e^-\rightarrow Fe}\qquad E^\circ=X$$

$$\Delta G_1^\circ=-2FX$$

$$\mathrm{Fe^{3+}+3e^-\rightarrow Fe}\qquad E^\circ=Y$$

$$\Delta G_2^\circ=-3FY$$

Required:

$$\mathrm{Fe^{3+}+e^-\rightarrow Fe^{2+}}\qquad E_3^\circ=?$$

Reverse the first reaction:

$$\mathrm{Fe\rightarrow Fe^{2+}+2e^-}$$

Adding this to the second reaction,

$$\mathrm{Fe^{3+}+3e^-\rightarrow Fe}$$

$$\mathrm{Fe\rightarrow Fe^{2+}+2e^-}$$

gives

$$\mathrm{Fe^{3+}+e^-\rightarrow Fe^{2+}}$$

Hence,

$$\Delta G_3^\circ=\Delta G_2^\circ-\Delta G_1^\circ$$

$$-FE_3^\circ=(-3FY)-(-2FX)$$

$$-FE_3^\circ=-3FY+2FX$$

Dividing by $$-F$$,

$$E_3^\circ=3Y-2X$$

Therefore,

$$E^\circ_{\mathrm{Fe^{3+}/Fe^{2+}}}=3Y-2X$$

Hence, the correct option is (B).