Correct Answer: $\frac{\sqrt{7}}{3}$

Solution : Given: $\sin \theta+\cos \theta=\frac{\sqrt{11}}{3}$    (equation 1)
Squaring both sides of the equation 1, we get,
$(\sin \theta+\cos \theta)^2=(\frac{\sqrt{11}}{3})^2$
⇒ $\sin^2 \theta+\cos^2 \theta+2\sin \theta \cos \theta=\frac{11}{9}$
⇒ $1+2\sin \theta \cos \theta=\frac{11}{9}$
⇒ $2\sin \theta \cos \theta=\frac{11}{9}-1$
⇒ $2\sin \theta \cos \theta=\frac{2}{9}$     (equation 2)
Substitute the value from the equation 2,
$(\sin \theta-\cos \theta)^2=\sin^2 \theta+\cos^2 \theta–2\sin \theta \cos \theta$
⇒ $(\sin \theta-\cos \theta)^2=1–\frac{2}{9}$
⇒ $(\sin \theta-\cos \theta)^2=\frac{7}{9}$
⇒ $(\sin \theta-\cos \theta)=\sqrt{\frac{7}{9}}$
$\therefore (\sin \theta–\cos \theta)=\frac{\sqrt7}{3}$
Hence, the correct answer is $\frac{\sqrt7}{3}$.