1. Relation for $g$

Given $T = 2\pi\sqrt{\frac{L}{g}}$

Squaring, $T^2 = 4\pi^2\frac{L}{g}$

So, $g = \frac{4\pi^2 L}{T^2}$

2. Error propagation

For $g \propto L\,T^{-2}$

The fractional error is $\frac{\Delta g}{g} = \frac{\Delta L}{L} + 2\frac{\Delta T}{T}$

3. Error in length measurement

Measured length: $L = 1.0\ \text{m}$

Minimum division of meter scale: $1\ \text{mm} = 0.001\ \text{m}$

So, $\Delta L = 0.001\ \text{m}$

Hence, $\frac{\Delta L}{L} = \frac{0.001}{1.0} = 0.001 = 0.1\%$

4. Error in time measurement

Measured time period: $T = 1.95\ \text{s}$

Stopwatch resolution: $\Delta T = 0.01\ \text{s}$

Thus, $\frac{\Delta T}{T} = \frac{0.01}{1.95} \approx 0.005128$

So, $2\frac{\Delta T}{T} \approx 2(0.005128) = 0.010256 = 1.0256\%$

5. Total percentage error in $g$

$\frac{\Delta g}{g} = 0.001 + 0.010256 = 0.011256$

In percentage, $0.011256 \times 100 \approx 1.13\%$

6. Option check

The percentage error is closest to: $\boxed{1.13\%}$

So the correct option is C.