$\text{ If the equation of the ellipse whose axes are coincident with the coordinate axes and }$

$\text{ which touches the straight lines }3x-2y-20=0\text{ and }x+6y-20=0\text{ is }\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,\text{ then a+b=}$

$\text{ Let the equation of the ellipse be }\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\dots \dots \dots \dots \dots .\left(1\right)$

Equation of the tangent to the ellipse in slope form is

$y=mx\pm \sqrt{a^2{m}^2+b^2}\dots \dots \dots \dots \dots \left(2\right)$

$\text{ Given equation of the tangent is }3x-2y-20=0$

Compare (2) and (3)

$\begin{array}{l}\Rightarrow m=\frac{3}{2}\text{ and }{a}^2{m}^2+b^2=100\\ \Rightarrow a^2\left(\frac{9}{4}\right)+b^2=100\\ \Rightarrow 9a^2+4b^2=400\dots \dots \dots \cdots \cdots \left(4\right)\\ \text{ Given equation of the another tangent is }x+6y-20=0\\ \Rightarrow y=-\frac{1}{6}x+\frac{10}{3}\dots \dots \dots \dots \dots \dots \left(5\right)\end{array}$

Compare (2) and (5)

$\begin{array}{l}\text{ We get }m=-\frac{1}{6}\text{ and }{a}^2{m}^2+b^2=\frac{100}{9}\\ \Rightarrow \frac{a^2}{36}+b^2=\frac{100}{9}\\ \Rightarrow a^2+36b^2=400\dots \dots \left(6\right)\end{array}$

Now solving equations (4) & (6) 

$\text{ we get }{a}^2=40\text{ and }{b}^2=10$

$\begin{array}{l}\Rightarrow a=\sqrt{40}=2\sqrt{10},b=\sqrt{10}\\ \Rightarrow a+b=3\sqrt{10}\end{array}$

Therefore, the correct answer is (3).