Continuous Uniform Distribution

The sample space is the square $$1\le X\le e,\qquad 1\le Y\le e$$

Hence, total area is $$=(e-1)^2$$

Now,

$$XY\le e$$ implies $$Y\le\frac{e}{X}$$

Therefore, favourable area is $$\int_1^e\left(\frac{e}{X}-1\right)dX$$

$$=\left[e\log X-X\right]_1^e$$

$$=(e-e)-(0-1)$$

$$=1$$

Hence,

$$\frac1k=\frac1{(e-1)^2}$$

So,

$$k=(e-1)^2$$

Using

$$e\approx2.718,$$

$$k\approx(1.718)^2$$

$$\approx2.952$$

Therefore,

$$100k\approx295.2$$

Hence,

$$\left\lfloor100k\right\rfloor=295$$