Solubility Product and Buffers
Ok, the answer is in three parts. Let's use $B^-$ to represent the benzoate anion. 

(1) Solubility of Silver Benzoate in pure Water

$\ce{AgB_{(s)} &lt;-&gt; Ag+_{(aq)} + B^−_{(aq)}} \quad\quad K_{sp} = 2.5 * 10^{-13}$

thus:
$\ce{[Ag^+] [B^-]} = 2.5 * 10^{-13}$

assume $\ce{[Ag^+] = [B^{-}]}$ then

$[Ag^+] = [B^-] = \sqrt{2.5 * 10^{-13}} \text{m/l}  = 5.0 * 10^{-7} \text{m/l}$

!! Check !!


  
    We want to check if $\ce{[B^{-}] &gt;&gt; [HB]}$ so that $[B^{−}_{(aq)}] + [HB_{(aq)}] \approx 5.0 * 10^{-7}$ and that the pH will stay at 7.
    
    We know that:
    
    (a) $\ce{HB_{(aq)} &lt;-&gt; H^{+}_{(aq)} + B−_{(aq)}}$
    
    (b) $\frac{[H+][B−]}{[HB]}=K_a=6.46∗10−5$
    
    (c) pH for pure water is 7
    
    So rearrange (b) and plug in our assumptions
    
    $ [HB] = \dfrac{[H+][B−]}{6.46∗10^{−5}} =  \dfrac{(1*10^{-7})(5.0*10^{-7})}{6.46∗10^{−5}} = 7.7 * 10^{-10}$
    
    Since $7.7 * 10^{-10} &lt;&lt; 5.0 * 10^{-7} $ we can safely ignore protonation of $\ce{B^-}$ to $\ce{HB}$. 
  


(2)Solubility of Silver Benzoate in a buffer of pH 3.19 

(a) $\frac{[H+][B−]}{[HB]}=K_a=6.46∗10^{−5}$

(b) pH for buffer is 3.19 therefore $\ce{[H^+]}$ = 6.45 * $10^{-4}$

(c) $\ce{AgB_{(s)} &lt;-&gt; Ag+_{(aq)} + B^−_{(aq)}} \quad\quad K_{sp} = 2.5 * 10^{-13}$

So rearrange (a) and plug in $\ce{[H^+]}$

$\dfrac{[B−]}{[HB]}= \dfrac{6.46∗10^{−5}}{6.45 * 10^{-4}} = 0.10 $

Thus in this acid solution we must consider the protonation of $\ce{B^{-}}$ to $\ce{HB}$ when solving for the solubility of silver benzoate. 

so we another equation.

(d) $\ce{AgB_{(s)} &lt;-&gt; Ag+_{(aq)} + B^{−}_{(aq)} + HB_{(aq)}}$

but we know that $\ce{[Ag+_{(aq)}] = [B^{−}_{(aq)}] + [HB_{(aq)}] = [B^{−}_{(aq)}] (1 + $\dfrac{\ce{[HB_{(aq)}]}}{\ce{[B^{−}_{(aq)}]}}$) = 11 [B^{−}_{(aq)}]}$

so from (c) we get:

$\ce{11[B^{−}]^2} = 2.5 * 10^{-13}$

$\ce{[B^{−}]}^2 = \dfrac{2.5 * 10^{-13}}{11} = 2.27 * 10^{-14}$

$\ce{[B^{−}]} = 1.51 * 10^{-7}$

and thus

$\ce{[HB]} = 10 \ce{[B^{−}]} = 1.51 * 10^{-6}$

$\ce{[Ag^+]} = 11 \ce{[B^{-}]} = 11 * 1.51 * 10^{-7} = 1.66 * 10^{-6}$

!! CHECK !!


  
    $[\ce{HB}] + \ce{[B^{-}]} = 1.51 * 10^{-6} + 1.51 * 10^{-7} = 1.66 * 10^{-6}$
    
    $[\ce{Ag^+}] [\ce{B^{-}}] = (1.66 * 10^{-6})(1.51 * 10^{-7}) = 2.51 * 10^{-13}$ 
  


(3) Ratio of Solubilities of Silver Benzoate

$\text{Ratio} = \dfrac{\text{solubility in buffer}}{\text{solubility in water}} = \dfrac{1.66 * 10^{-6}}{5.0 * 10^{-7}} = 3.3 $