Idea
We need to check which metals can be oxidized by $\mathrm{NO_3^-}$ in aqueous solution.
This means $\mathrm{NO_3^-}$ will act as the oxidizing agent and itself get reduced: $\mathrm{NO_3^-}+4\mathrm{H}^+ +3e^- \rightarrow \mathrm{NO}(g)+2\mathrm{H_2O},\qquad E^\circ=0.97\,\text{V}$

Condition for spontaneity
If a metal $M$ is oxidized, then the corresponding reduction half-reaction is given in the table.
For the cell reaction: $E^\circ_{\text{cell}}=E^\circ_{\text{cathode}}-E^\circ_{\text{anode}}$ Here cathode is nitrate reduction: $E^\circ_{\text{cathode}}=0.97\,\text{V}$ If $E^\circ_{\text{cell}}>0$, the metal will be oxidized by $\mathrm{NO_3^-}$.

Check each metal

(a) Vanadium

Given: $\mathrm{V}^{2+}+2e^- \rightarrow \mathrm{V},\qquad E^\circ=-1.19\,\text{V}$ So, $E^\circ_{\text{cell}}=0.97-(-1.19)=2.16\,\text{V}>0$ Hence, $\mathrm{V}$ will be oxidized.

(b) Iron

Given: $\mathrm{Fe}^{3+}+3e^- \rightarrow \mathrm{Fe},\qquad E^\circ=-0.04\,\text{V}$ So, $E^\circ_{\text{cell}}=0.97-(-0.04)=1.01\,\text{V}>0$ Hence, $\mathrm{Fe}$ will be oxidized.

(c) Silver

Given: $\mathrm{Ag}^+ + e^- \rightarrow \mathrm{Ag},\qquad E^\circ=0.80\,\text{V}$ So, $E^\circ_{\text{cell}}=0.97-0.80=0.17\,\text{V}>0$ Hence, $\mathrm{Ag}$ will be oxidized.

(d) Gold

Given: $\mathrm{Au}^{3+}+3e^- \rightarrow \mathrm{Au},\qquad E^\circ=1.40\,\text{V}$ So, $E^\circ_{\text{cell}}=0.97-1.40=-0.43\,\text{V}<0$ Hence, $\mathrm{Au}$ will not be oxidized.

Count the metals oxidized
The metals oxidized by $\mathrm{NO_3^-}$ are: $\mathrm{V},\ \mathrm{Fe},\ \mathrm{Ag}$ Total number of metals $=3$.

Final Answer $\boxed{3}$

Comparison with stored answer
Stored correct answer = $3$
Derived answer = $3$
Hence, they agree.