= Solution
All four <angle> measurements have the same <standard uncertainty>, so <ordinary least squares> minimizes
$$
S(c)=\sum_{i=1}^4(y_i-cx_i)^2.
$$
<Differentiation> gives
$$
\frac{dS}{dc}=-2\sum_i x_i(y_i-cx_i)=0,
\qquad \frac{d^2S}{dc^2}=2\sum_i x_i^2>0.
$$
Thus the unique minimum is
$$
\boxed{\hat c=\frac{\sum_i x_iy_i}{\sum_i x_i^2}
=\frac{\overline{xy}}{\overline{xx}}.}
$$
The overbars denote <sample means> over the four observations. Numerically,
$$
\sum_i x_iy_i=542.888740\,\mathrm{m^{-2}},\qquad
\sum_i x_i^2=3.00072844\times10^7\,\mathrm{m^{-4}},
$$
so
$$
\boxed{\hat c=1.80919\times10^{-5}\,\mathrm{m^2},\qquad
\hat G=\frac{\kappa\hat c}{mML}=9.59\times10^{-11}\,\mathrm{m^3\,kg^{-1}\,s^{-2}}.}
$$
The fixed $x_i$ have negligible uncertainty. Assuming <independent> errors in the four $y_i$, <propagation of uncertainty> in quadrature gives
$$
\sigma_{\hat c}^2=\sum_i\left(\frac{\partial\hat c}{\partial y_i}\right)^2\sigma_y^2
=\frac{\sigma_y^2\sum_i x_i^2}{(\sum_i x_i^2)^2}
=\frac{\sigma_y^2}{\sum_i x_i^2}.
$$
Hence
$$
\boxed{\sigma_{\hat c}=3.65\times10^{-7}\,\mathrm{m^2},\qquad
\hat c=(1.809\pm0.037)\times10^{-5}\,\mathrm{m^2}.}
$$
If the period measurement is <independent> of the <angle> measurements, including its calibration uncertainty gives
$$
\frac{\sigma_G}{\hat G}=\sqrt{\left(\frac{\sigma_{\hat c}}{\hat c}\right)^2+
\left(\frac{\sigma_\kappa}{\kappa}\right)^2}=0.0314,
\qquad \hat G=(9.59\pm0.30)\times10^{-11}\,\mathrm{m^3\,kg^{-1}\,s^{-2}}.
$$
This last uncertainty includes the stated random measurement errors only. Whether the model describes the data must be checked before treating it as a reliable measurement of the <gravitational constant>.
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