Finding the frequency factor when raising temperature
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The Arrhenius equation is:

$$ k = A e^{-\frac{E_a}{RT}}$$

A linearized form of the Arrhenius equation is 

$$ \ln{k} = \ln{A} - \frac{E_a}{R}T^{-1}$$

This equation linearly relates $\ln{k}$ to $T^{-1}$: the intercept is $\ln{A}$ and the slope is $-\frac{E_a}{R}$.  

To fully define a line, we need two parameters.  This can be two completely specified points that lie on the line, or any single point on the line plus a slope for the line.  For this problem that would mean either (a) two temperatures and two rates, or (b) one temperature, one rate, and one slope.

Using the information we are given:

$$ \ln{k} = \ln{A} - \frac{E_a}{R}T_1^{-1}$$
$$ \ln{2k} = \ln{2} + \ln{k} = \ln{A} - \frac{E_a}{R}T_2^{-1}$$

Any way we combine those two equations will only yield an equation equivalent to

$$\ln{2} = -\frac{E_a}{R}\left(T_2^{-1} - T_1^{-1} \right)$$

in which $\ln{k}$ and $\ln{A}$ have both cancelled out.  That is because the starting two linear equations have the same coefficients for $\ln{k}$ and $\ln{A}$ in each equation.  Similarly, the two equations $2x=y$ and $2x+2=y+2$  can't be solved for $x$ and $y$.

The problem as stated gives us only a slope, but not even a single point that lies on the line.  The rate could double by going from 1,000,000 $\text{s}^{-1}$ to 2,000,000 $\text{s}^{-1}$ (a very fast reaction!) or by going from 0.1 $\text{yr}^{-1}$ to 0.2 $\text{yr}^{-1}$ (pretty slow).  There is no way to find the intercept of a line when we are given only the slope.  Thus, there is no way to solve for $A$ using the information given.