1. Write the cell reaction

Given cell: $\text{Zn}|\text{Zn}^{2+}(0.1\,M)||\text{Ag}^+(0.01\,M)|\text{Ag}$

At anode: $\text{Zn} \rightarrow \text{Zn}^{2+}+2e^-$

At cathode: $\text{Ag}^+ + e^- \rightarrow \text{Ag}$

Balancing electrons: $2\text{Ag}^+ + 2e^- \rightarrow 2\text{Ag}$

Overall reaction: $\text{Zn} + 2\text{Ag}^+ \rightarrow \text{Zn}^{2+} + 2\text{Ag}$

So, number of electrons transferred is $n=2$

2. Calculate standard emf of the cell

$E^\circ_{\text{cell}}=E^\circ_{\text{cathode}}-E^\circ_{\text{anode}}$

$E^\circ_{\text{cell}}=0.80-(-0.76)=1.56\,V$

3. Write the reaction quotient

For the overall reaction, $Q=\frac{[\text{Zn}^{2+}]}{[\text{Ag}^+]^2}$

Given: $[\text{Zn}^{2+}]=0.1,\qquad [\text{Ag}^+]=0.01$

Hence, $Q=\frac{0.1}{(0.01)^2}=\frac{0.1}{10^{-4}}=1000$

4. Apply Nernst equation at 298 K

$E_{\text{cell}}=E^\circ_{\text{cell}}-\frac{0.059}{n}\log Q$

Substitute values: $E_{\text{cell}}=1.56-\frac{0.059}{2}\log(1000)$

Since, $\log(1000)=3$

So, $E_{\text{cell}}=1.56-\frac{0.059}{2}\times 3$

$E_{\text{cell}}=1.56-0.0885=1.4715\,V$

5. Express in the form $x \times 10^{-2}$ V

$1.4715\,V = 147.15 \times 10^{-2}\,V$

Rounded to nearest integer, $x=147$

Hence, the required answer is: $\boxed{147}$