= Solution
Evaluate the source's printed logarithm using the numerical magnitude value 21.0, with base-ten <logarithms>:
$$
\log_{10}\frac{R_e}{\mathrm{kpc}}=1.3\log_{10}(250)-0.8\log_{10}(21)-1.36=0.69955,
\qquad \boxed{R_e\simeq5.01\,\mathrm{kpc}.}
$$
This is the literal printed-fit convention identified above. Convert the mean visual <surface brightness> to a linear <luminosity> per projected <area>. At 10 <parsecs>, one square <arcsecond> subtends $100/(206264.806)^2\,\mathrm{pc^2}$. The definition of <absolute magnitude> consequently gives
$$
\frac{\mathcal I_e}{L_{V,\odot}\,\mathrm{pc^{-2}}}=10^{-0.4(\mu_e-M_{V,\odot}-21.572)}
=10^{-0.4(21.0-4.8-21.572)}\simeq140.9.
$$
The constant $21.572=5\log_{10}(206264.806)-5$ is the conversion between the angular and physical projected <area>. Since the <effective radius> encloses half the total <luminosity>,
$$
\boxed{L_V=2\pi R_e^2\mathcal I_e\simeq2.22\times10^{10}L_{V,\odot}.}
$$
For the stipulated <stellar population> of <Solar-type stars>, approximate the stellar <mass-to-light ratio> by $M_\odot/L_{V,\odot}$. Summing 100 such galaxies then gives
$$
\boxed{M_{*,\mathrm{cluster}}\simeq100\frac{L_V}{L_{V,\odot}}M_\odot=2.22\times10^{12}M_\odot.}
$$
This assumes negligible <interstellar extinction> and that the supplied <surface brightness> is an intrinsic or already corrected value; an observed cosmological <surface brightness> requires additional dimming and passband corrections before this conversion.
For comparison only, if the empirical coefficients and intercept were instead meant for $\log_{10}(\mathcal I_e/[L_{V,\odot}\,\mathrm{pc^{-2}}])$, substituting $140.9$ in that fit would give $R_e=1.09\,\mathrm{kpc}$, $L_V=1.06\times10^9L_{V,\odot}$, and $M_{*,\mathrm{cluster}}=1.06\times10^{11}M_\odot$. Those are conditional alternative results, not an unannounced correction of the printed units.
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