The time period of a simple harmonic oscillator is T=2 π (k)/(m) ....

The given formula for time period is
$T = 2\pi \sqrt{\frac{k}{m}}$.
Squaring both sides, we get $T^2 = 4\pi^2 \frac{k}{m}$, which gives $k = \frac{4\pi^2 m}{T^2}$.
The relative error in $k$ is given by $\frac{\Delta k}{k} = \frac{\Delta m}{m} + 2\frac{\Delta T}{T}$.
Given: $m = 10$ g, $\Delta m = 10$ mg $= 0.01$ g.
Total time for $n=50$ oscillations is $t = 60$ s with resolution $\Delta t = 2$ s.
Since $T = t/n$, the relative error in $T$ is the same as in $t$: $\frac{\Delta T}{T} = \frac{\Delta t}{t}$.
Substituting the values: $\frac{\Delta k}{k} = \frac{0.01}{10} + 2\left(\frac{2}{60}\right)$.
$\frac{\Delta k}{k} = 0.001 + \frac{2}{30} = 0.001 + 0.0666...$
Percentage error $= \left(0.001 + \frac{1}{15}\right) \times 100 = 0.1 + \frac{100}{15} = 0.1 + 6.666... = 6.766...\%$.
Rounding to two decimal places, we get $6.76\%$.