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[tex] \Large \mathbb{SOLUTION:} [/tex]
[tex] \begin{array}{l} \text{Let }\frak{t}\text{ be the time in hour/s, after }\frak{8:00},\text{the man} \\ \text{should arrive in his office.} \\ \\ \text{We know that }\bold{Distance} = \bold{Rate}\times \bold{Time}. \\ \\ \text{Solving for }\frak{t}, \\ \\ \begin{aligned} \quad \quad \frak {2\left(t - \frac{3}{60}\right)} &= \frak{3\left(t - \frac{30}{60} + \frac{6}{60}\right)} \\ \frak{2t - \frac{1}{10}} &= \frak{3t - \frac{6}{5}} \\ \frak{\frac{6}{5} - \frac{1}{10}} &= \frak{3t - 2t} \\ \frak{t} &= \frak{\frac{11}{10}\ \text{hrs} = 66}\ \text{mins} \end{aligned} \\ \quad\quad \implies \frak{8:00 + 66}\ \text{mins} = \boxed{\frak{9:06}} \\ \\ \text{Therefore, the man should arrive in his office} \\ \text{at }\frak{9:06}. \end{array} [/tex]