Maybe this problem solve like this
Look at the Sec. 4.6 (Systematic Errors)
In part (c) Problem 4.28 We must find the systematic error of length of the pendulum ($\delta l_{sys}$). Given that the value of $\delta g_{tot}$ is $|g_{accepted} - \bar{g}|$.
Using the above argument and Eq. 4.26 $\left[ \delta g_{tot} = \sqrt{\delta g_{sys}^2 + \delta g_{ran}^2} \,\,\right]$, $$ \begin{align} \delta g_{tot} &= \sqrt{\delta g_{sys}^2 + \delta g_{ran}^2} \\ |g_{accepted} - \bar{g}| &= \sqrt{\delta g_{sys}^2 + \sigma_{\bar{g}}^{\,\,\,2}} \\ \delta g_{sys} &= \sqrt{\mathrm{discrepancy}^2 - \sigma_{\bar{g}}^{\,\,\,2}} \end{align} $$
and propagation error techniques for the equation $g = 4\pi l/T^2$ gives the formula for $\delta g_{sys}$ $$ \begin{align} \frac{\delta g_{sys}}{\bar{g}} = \sqrt{\left(\frac{\delta l_{sys}}{\bar{l}}\right)^2 + \left(2\frac{\delta{T_{sys}}}{\bar{T}}\right)^2} \end{align} $$
Since there was no problem with the measurement of the period $T$ ($\delta T_{sys} = 0$), we have the propagation error for $\delta g_{sys}$ $$ \begin{align} \frac{\delta g_{sys}}{\bar{g}} = \frac{\delta l_{sys}}{\bar{l}} \end{align} $$
Therefore, the systematic of the length $l$ (in percentage unit) is $$ \frac{\delta l_{sys}}{\bar{l}} = \frac{\sqrt{\mathrm{discrepancy}^2 - \sigma_{\bar{g}}}}{\bar{g}} \cdot 100\% $$
Substituting all the value has given and obtained from the previous question, we have the final result
$$
\begin{align}
\frac{\delta l_{sys}}{\bar{l}} &= 1.43793098102868 \% \\
&\approx 1.4\%
\end{align}
$$
$$ \begin{align} \mathfrak{God\,Bless\,You} \end{align} $$
