Some important formulae to be kept in mind before solving the question:
The Standard equation of ellipse is given by : $\frac{x^{2}}{a^{2}}$ $+$ $\frac{y^{2}}{b^{2}}$ = 1
The distance of the focus of the ellipse from the centre $\left ( 0,0 \right )$ can be found using $c=\pm \sqrt{a^{2}-b^{2}}$
The length of the latus rectum(the chord passing through the focus perpendicularly) of the ellipse is : $2\times \frac{b^{2}}{a}$
Now, coming back to the question, let the intesection points of the tangent with the $x$-$axis$ and $y$-$axis$ in the first quadrant
be $\left ( d,0 \right )$ and $\left ( 0,c \right )$ respectively.
Also the co-ordinates of the focus in the first quadrant would be $c=\pm \sqrt{9-5}$ = $\pm 2$ i.e. $\left ( 2,0 \right )$
The ends of latus rectum in the first quadrant will be $\left ( 2,\frac{5}{3} \right )$ . So, now the equation of tangent to the ellipse at point $\left ( x_{1},y_{1} \right )$ is given by :
$\left ( b^{2}x_{1} \right )x$ + $\left ( a^{2}y_{1} \right )y$ $=$ $a^{2}b^{2}$ (Source: Tangent to ellipse)
Using the above equation, the equation of tangent would be: $\left ( 5\times 2 \right )x$ $+$ $\left ( 9\times \frac{5}{3} \right ) y$ = $45$ i.e. $2x+3y=9$
The points $\left ( d,0 \right )$ and $\left ( 0,c \right )$ satisfy the above equation.
So, $2\times d=9 \Rightarrow d=\frac{9}{2}$ and $3\times c=9\Rightarrow c=3$.
Now that we have got the end points of the tangents in the first quadrant, we know that area of the whole quadrilateral would be :
$4\times$ area of the triangle in the first quadrant (by symmetry)
Area of the triangle formed by the tangent with $x$-$axis$ and $y$-$axis$ in the first quadrant is: $\frac{1}{2}\times \frac{9}{2}\times3$ = $\frac{27}{4}$.
$\therefore$ Area of the quadrilateral is : $4\times \frac{27}{4}=27$ $sq.$ $units$
Option (A) is the correct answer.