No, a quick calculation yields a $\Delta v$ of about 4.6 km/s and you need about 9 km/s to get to low-Earth orbit. You'll lose a lot of that velocity to aerodynamic drag as well as the vertical portion of the flight, so a rough estimate of the final speed at burnout would be somewhere around 3 km/s -- or about 10,000 km/h.
Using the specifications from the SRB Wikipedia article (which admittedly have some conflicts):
$m_{full} = 590000 kg$
$m_{empty} = 91000 kg$
$I_{SP} = 250 s$ (242 at sea-level, 268 in a vacuum, most of the flight is in the atmosphere so we'll round up to 250)
$\Delta v = g_0 I_{SP}\ln \frac{m_{full}}{m_{empty}}$
Here is a simulation of a launch composed of vertical flight until reaching 100 m altitude, then pitching over to a constant pitch of 85 deg until reaching 1000 m altitude, then reverting to a gravity turn at 0 deg angle-of-attack. This sends the SRB to an altitude over 500 km, with a peak speed around 3.7 km/s at burnout.


Here is a look at the pitch, flight-path angle (direction of motion), and angle-of-attack (difference between pitch and flight-path angle) as a function of altitude. Although it might seem strange that the rocket starts pitching over after only 100 m of flight, this is necessary to start gaining enough velocity to reach orbit -- the rocket is still climbing throughout most of the flight.

Note that I'm using a simple atmospheric model for the air density:
$\rho = 1.225 e ^\frac{h}{8000}$
And a constant drag coefficient of $C_D = 0.8$:
$D = \frac{1}{2} \rho v^2 C_D A$
Mass flow rate is assumed to be constant, computed from the stated sea-level thrust (12 MN) and specific impulse (242 s), whereas specific impulse is computed based on a simple mixture rule:
$I_{SP} = I_{SP,sea} \frac{\rho}{\rho_{sea}} + I_{SP,vac}\left( 1 - \frac{\rho}{\rho_{sea}}\right)$
Then thrust is computed from the specific impulse and mass-flow:
$T = \dot{m} g_0 I_{SP}$
Note that the equations of motion in 2D are:
$\ddot{r} = -\frac{\mu}{r^2} + r \dot{\theta}^2 + \frac{T}{m} \sin \phi - \frac{D}{m} \sin \gamma$
$\ddot{\theta} = -\frac{2 \dot{r} \dot{\theta}}{r} + \frac{T}{m r} \cos \phi - \frac{D}{m r} \cos \gamma$
Where $r$ and $\theta$ are the polar coordinates about the Earth's centre, $m$ is the vehicle mass, $\phi$ is the vehicle pitch and $\gamma$ is the flight-path angle (velocity direction) -- both of those angles are with respect to the local horizon.
As for all-solid fuel launch vehicles, there are a number of them that have been and are still in use (such as the Minotaur family). But, as LocalFluff pointed out, they are usually only sensible for smaller payloads. Lunar Prospector was launched to the moon on an all-solid Athena II although it looks like the upper stage used hydrazine (not solid, but not a complex liquid motor either).