Laplace and KVL

I was preparing for my exam by solving questions and there is this one question I was stuck on but when I saw its solution provided by book, I think KVL applied is not right or I am not able to understand it properly. Can some please help.

Books says - (Xc + 6K + 4K)I(s) = - 160/9s

But I think it should be (Xc + 6K + 4K) I(s) = 160/9s

Here is the question. It all depends on the direction you choose for the current. Book's answer is using $$\I_c\$$, that is, counter-clockwise direction, so the voltage across the source will be seen like a voltage drop. You are trying to use $$\i_r\$$, which is clockwise direction, so the source produces a voltage rise, opposite to the drop over the impedances. That is why the book's answer has the negative sign on the $$\160/9s\$$ term.
In your case, you will obviously arrive to $$\i_r=\frac{16}{9}e^{-100t/3}\$$, and when you calculate $$\V_c(t)\$$, you will have to use this equation: $$V_c(t)=V_{C10}-\frac{1}{c}\int_0^t{i_r(t)dt}$$ with the minus sign, because the voltage source and the impedance have opposite sign for the current you chose.
If you replace the value for $$\i_r\$$, you will get the same result as the book.