1. For a simple pendulum,

$T = 2\pi\sqrt{\frac{l}{g}} \quad \Rightarrow \quad g = \frac{4\pi^2 l}{T^2}$

Hence, the fractional error in $g$ is

$\frac{\Delta g}{g} = \frac{\Delta l}{l} + 2\frac{\Delta T}{T}$

So percentage error is

$\%\text{ error in } g = \left(\frac{\Delta l}{l} + 2\frac{\Delta T}{T}\right)\times 100$

2. Error in length measurement:

Least count of length $=0.1\,\text{cm}$, so

$\Delta l = 0.1\,\text{cm}$

3. Error in time period:

Since $T = \dfrac{t}{n}$, where $t$ is total time for $n$ oscillations,

$\frac{\Delta T}{T} = \frac{\Delta t}{t}$

because $n$ is counted exactly.

Given least count of time $=0.1\,\text{s}$,

$\Delta t = 0.1\,\text{s}$

Now calculate for each student.

Student 1

Given:

$l=64.0\,\text{cm}, \quad t=128.0\,\text{s}, \quad n=8, \quad T=16.0\,\text{s}$

Fractional error:

$\frac{\Delta g}{g} = \frac{0.1}{64.0} + 2\cdot \frac{0.1}{128.0}$

$= 0.0015625 + 0.0015625 = 0.003125$

Percentage error:

$E_1 = 0.3125\%$

Student 2

Given:

$l=64.0\,\text{cm}, \quad t=64.0\,\text{s}, \quad n=4, \quad T=16.0\,\text{s}$

Fractional error:

$\frac{\Delta g}{g} = \frac{0.1}{64.0} + 2\cdot \frac{0.1}{64.0}$

$= 0.0015625 + 0.003125 = 0.0046875$

Percentage error:

$E_2 = 0.46875\%$

Student 3

Given:

$l=20.0\,\text{cm}, \quad t=36.0\,\text{s}, \quad n=4, \quad T=9.0\,\text{s}$

Fractional error:

$\frac{\Delta g}{g} = \frac{0.1}{20.0} + 2\cdot \frac{0.1}{36.0}$

$= 0.005 + 0.005555\ldots = 0.010555\ldots$

Percentage error:

$E_3 \approx 1.0556\%$

4. Compare the errors:

$E_1 = 0.3125\%, \quad E_2 = 0.46875\%, \quad E_3 \approx 1.0556\%$

Thus, the minimum percentage error is obtained by student 1.

$\boxed{1}$