Answer
The mass of the portion of the Milky Way inside the sun's orbit:
$M \simeq 2.623 \times {10^{41}}$ kg
Work Step by Step
We have the radius of the orbit $a \simeq 2.8 \times {10^{17}}$ km and velocity $v \simeq 250$ km/s. So, the period of the orbit is
$T = \frac{{2\pi a}}{v} = \frac{{2\pi \times 2.8 \times {{10}^{17}}}}{{250}} = 7.037 \times {10^{15}}$ s
From Exercise 2 we obtain the mass of the Milky Way galaxy:
$M = \left( {\frac{{4{\pi ^2}}}{G}} \right)\left( {\frac{{{a^3}}}{{{T^2}}}} \right)$,
where $G$ is the gravitational constant,
$G \simeq 6.673 \times {10^{ - 11}}$ ${m^3}k{g^{ - 1}}{s^{ - 2}}$
So,
$M = \left( {\frac{{4{\pi ^2}}}{{6.673 \times {{10}^{ - 11}}}}} \right)\frac{{{{\left( {2.8 \times {{10}^{20}}} \right)}^3}}}{{{{\left( {7.037 \times {{10}^{15}}} \right)}^2}}} \simeq 2.623 \times {10^{41}}$ kg
