The area of the region
$$\left\{(x, y) : 0 \leq x \leq \frac{9}{4}, 0 \leq y \leq 1, x \geq 3y, x + y \geq 2\right\}$$
is

The region is described by all $$(x,y)$$ that simultaneously satisfy

$$0 \le x \le \frac94,\; 0 \le y \le 1,\; y \le \frac{x}{3},\; y \ge 2-x$$

The two slant lines $$y=\frac{x}{3}$$ (slope $$\frac13$$) and $$y=2-x$$ (slope $$-1$$) meet where

$$2-x=\frac{x}{3} \;\Longrightarrow\; 6-3x=x \;\Longrightarrow\; x=\frac32,\; y=\frac12$$

Hence the feasible $$x$$-interval starts at $$x=\frac32$$. For each $$x$$ we now determine the lower and upper $$y$$-bounds.

Here $$2-x \gt 0$$, so the lower bound is $$y_{\min}=2-x$$. The upper bound is always $$y_{\max}=x/3$$ because $$x/3 \le 1$$ in this range. The vertical strip height is

$$h_1(x)=\frac{x}{3}-(2-x)=\frac{4x}{3}-2$$

Area in this case

$$A_1=\int_{3/2}^{2}\left(\frac{4x}{3}-2\right)dx =\left[\frac{2}{3}x^{\,2}-2x\right]_{3/2}^{2} =-\frac43+\frac32=\frac16$$

Now $$2-x \le 0$$, but $$y \ge 0$$ from the box constraint, so $$y_{\min}=0$$. Again $$y_{\max}=x/3$$. Strip height:

$$h_2(x)=\frac{x}{3}$$

Area in this case

$$A_2=\int_{2}^{9/4}\frac{x}{3}\,dx =\left[\frac{x^{2}}{6}\right]_{2}^{9/4} =\frac{1}{6}\left(\frac{81}{16}-4\right) =\frac{17}{96}$$

Total area

$$A=A_1+A_2=\frac16+\frac{17}{96} =\frac{16}{96}+\frac{17}{96} =\frac{33}{96} =\frac{11}{32}$$

Hence the required area is $$\dfrac{11}{32}$$.

Option A which is: $$\frac{11}{32}$$