Find the emf of the cell in which the following reaction takes place at $298 \mathrm{K}$

$\mathrm{Ni}\left(\mathrm{s}\right)+2{\mathrm{Ag}}^{+}(0.001 \mathrm{M})\to {\mathrm{Ni}}^{2+}(0.001\mathrm{M})+2\mathrm{Ag}\left(\mathrm{s}\right)$

(Given that $\mathrm{E}{°}_{\mathrm{cell}}=1.05 \mathrm{V}, \frac{2.303 \mathrm{RT}}{\mathrm{F}}=0.059$ at $298 \mathrm{K}$)

Given reaction: $\mathrm{Ni}\left(\mathrm{s}\right)+2{\mathrm{Ag}}^{+}(0.001 \mathrm{M})\to {\mathrm{Ni}}^{2+}(0.001\mathrm{M})+2\mathrm{Ag}\left(\mathrm{s}\right)$

Applying Nernst equation we have:
${\mathrm{E}}_{\left(\mathrm{cell}\right)}={\mathrm{E}}_{\left(\mathrm{cell}\right)}^0-\frac{2.303 \mathrm{RT}}{\mathrm{nF}}\log \frac{\left[{\mathrm{Ni}}^{2+}\right]\left[\mathrm{Ag}\right]}{{\left[{\mathrm{Ag}}^{+}\right]}^2\left[\mathrm{Ni}\right]}$
Active mass of solid is taken to be unity so $\left[\mathrm{Ni}\left(\mathrm{s}\right)\right]=\left[\mathrm{Ag}\left(\mathrm{s}\right)\right]=1$
${\mathrm{E}}_{\left(\mathrm{c}\mathrm{e}\mathrm{l}\mathrm{l}\right)}={\mathrm{E}}_{\left(\mathrm{c}\mathrm{e}\mathrm{l}\mathrm{l}\right)}^0-\frac{0.059}{\mathrm{n}}\log \frac{\left[\mathrm{N}{\mathrm{i}}^{2+}\right]}{{\left[\mathrm{A}{\mathrm{g}}^{+}\right]}^2}$
$=1.05-\frac{0.0591}{2}\log \frac{\left(0.001\right)}{{\left(0.001\right)}^2}$
$=1.05-0.0295 \log (1\times {10}^3)$
$=1.05-0.0295 \times 3$
$=0.9615 \mathrm{V}$

Therefore, the emf of the cell is $0.9615 \mathrm{V}$. 

Note: Question is modified little bit for academic accuracy. $\mathrm{E}{°}_{\mathrm{cell}}=1.05 \mathrm{V} \mathrm{in} \mathrm{place} \mathrm{of} \mathrm{E}{°}_{\mathrm{cell}}=10.5 \mathrm{V}$ as given in NEET 2022 paper.