The density of ice at $0^{\circ}\mathrm{C}$ is 0.915 gram/cc and that of liquid water at $0^{\circ}\mathrm{C}$ is 0.99987 gram/cc. The work done for melting 1 mole of ice at 1.00 bar (assuming work is done only due to expansion) is approximately

For melting at constant pressure, the expansion work is $$w=-P\Delta V.$$ We need the volume change for $1$ mole, whose mass is $18$ gram. Volume of ice: $$V_{\text{ice}}=\frac{18}{0.915}\,\text{cc}\approx 19.67\,\text{cc}.$$ Volume of water: $$V_{\text{water}}=\frac{18}{0.99987}\,\text{cc}\approx 18.00\,\text{cc}.$$ Hence, $$\Delta V=V_{\text{water}}-V_{\text{ice}}\approx -1.67\,\text{cc}=-1.67\times 10^{-6}\,\text{m}^3.$$ At $1.00$ bar, $$P=10^5\,\text{Pa}.$$ Therefore, $$w=-P\Delta V=-(10^5)(-1.67\times 10^{-6})\,\text{J}\approx 0.17\,\text{J}.$$ So the magnitude of work done is approximately $0.17\,\text{J}$, which matches option $\text{A}$.