Given $T = 2\pi \sqrt {\large\frac{L}{g}}$ where $T = \large\frac{t}{n}$ where $t$ is the time taken for $n$, the number of oscillations,

$\large\frac{t}{n}$$ = 2\pi \sqrt {\large\frac{L}{g}}$$ \rightarrow$ (Squaring and solving for $g$) $g = \large\frac{4\pi^2Ln^2}{t^2}$

Substituting, we get $g = \large\frac{4 \times \pi^2 \times 0.2 \times 100^2}{90^2}$$ = 9.74\;m\;s^{-2}$

Given $g = \large\frac{4\pi^2Ln^2}{t^2}$, the relative error in $g$ is:

$\large\frac{\Delta g}{g}$$ = \large\frac{\Delta L}{L}$$ + \large\frac{2\Delta t}{t}$

$\Rightarrow$ $\large\frac{\Delta g}{g}$$ = \large\frac{0.1}{20} $$ + \large\frac{2\times 1}{90}$$ = 0.005+0.022 = 0.027$

$\Rightarrow \Delta g = 0.027 \times 9.74 = 0.26\;m\;s^{-2}$

Rounding off the first significant digit, we get $\Delta g = 0.3 \;m\;s^{-2}$

Therefore, $g$ must also be rounded off as $9.7 \;m\;s^{-2}$.

Therefore, $g = (9.7 \pm 0.3) \;m\;s^{-2}$