Let $X$ be the oxidation number of cobalt. $X+(-2)+(-1)=0$ or $X=3 .$ Hence, the oxidation number is 3 .

The outer electronic configuration of cobalt (atomic number 27 ) is $3 d^{7} 4 s^{2} .$ The outer electronic configuration of $Co ^{+3}$ will be $3 d^{6} 4 s^{0} .$ Thus it contains six d electrons.

$\left[ Co \left( NH _{3}\right)_{5} CO _{3}\right] ClO _{4}$ is inner orbital or low spin complexes. all the electrons are paired due to $d^{2} s p^{3}$ hybridization of $C o^{+3}$ ion. Hence, the number of unpaired d electrons is zero.