Let us assume $$\sqrt{13+3\sqrt{\frac{23}{3}}} +\sqrt{13-3\sqrt{\frac{23}{3}}}=x $$ Squaring and simplifying gives $$26+20=x^2$$ which gives $$x=+\sqrt{46}$$
Since $6= \sqrt{36}<\sqrt{46}<\sqrt{49}=7$
The result is $(D)$
Let us assume $$\sqrt{13+3\sqrt{\frac{23}{3}}} +\sqrt{13-3\sqrt{\frac{23}{3}}}=x $$ Squaring and simplifying gives $$26+20=x^2$$ which gives $$x=+\sqrt{46}$$
Since $6= \sqrt{36}<\sqrt{46}<\sqrt{49}=7$
The result is $(D)$