Walking at $75\%$ of his usual speed, Raghu is $80$ minutes late to his office. What is his usual time in hours to cover the same distance?

Since the distance to the office is constant, speed is inversely proportional to time ($S\propto \frac{1}{T}$).Let the usual speed be $S$ and usual time be $T$.New speed = $75\%\text{ of }S=\frac{3}{4}S$ The ratio of usual speed to new speed is:
$$\text{Ratio of Speed}=1:\frac{3}{4}=4:3$$
Since time is inversely proportional to speed, the ratio of usual time to new time will be:
$$\text{Ratio of Time}=3:4$$
Let the usual time be $3x$ and the new time be $4x$.The difference in time is given as $80$ minutes:
$$4x-3x=80⟹x=80\text{ minutes}$$
Therefore, the usual time is:
$$\text{Usual Time}=3x=3\times 80=240\text{ minutes}$$
Converting minutes into hours:
$$\text{Usual Time in hours}=\frac{240}{60}=4\text{ hours}$$