Astronomers use the following relationship to estimate the mass $M$ of a galaxy $$M = \frac{v^2 r}{G}$$ where $v$ is the orbital speed of a star in the outer regions of the galaxy, in m s$^{-1}$, $r$ is the orbital radius of the star, in m, $G$ is the Universal Constant of Gravitation - Scottish Highers Physics - Question 25 - 2022

First, convert the orbital speed from km/s to m/s:

$v = 120 \text{ km/s} = 120,000 \text{ m/s}$

Using the radius provided:

$r = 4.0 \times 10^6 \text{ m}$

We can substitute these values into the formula:

$M = \frac{v^2 r}{G}$

To proceed, we need the value for $G$, which is approximately:

$G = 6.674 \times 10^{-11} \text{ m}^3 \text{ kg}^{-1} \text{s}^{-2}$

Now, calculate $v^2$:

$v^2 = (120,000)^2 = 14,400,000,000 \text{ m}^2/\text{s}^2$

Now calculate the mass:

$M = \frac{14,400,000,000 \times (4.0 \times 10^6)}{6.674 \times 10^{-11}}$

Calculating this will yield:

$M \approx 8.6 \times 10^{10} \text{ kg}$

Thus, the mass of the Triangulum galaxy is approximately $8.6 \times 10^{10}$ kg, which corresponds to option E.