Question regarding slew-rate
There are two things at play here:

1) Opamp bandwidth

2) Opamp slew rate

Let's say your opamp has the following transfer function (a low pass filter):

$$ H(s)=\dfrac{10}{\frac{s}{\omega_c}+1}$$

So, at dc the gain is 10 and the cutoff frequency is \$\omega_c\$. 

The response of the circuit to a unit step input (just considering one half of the square wave) is:

$$ v_o=10(1-e^{-\omega_ct})$$

This is just a signal that will increase exponentially at the beginning before reaching steady state.

Let's check if the output is going to be BW-limited or SR-limited.

$$\dfrac{dv_o}{dt}= 10\omega_ce^{-\omega_ct}$$

The slope is the highest near zero, so the initial slope is:

$$\dfrac{dv_o}{dt}\bigg|_{t=0}= 10\omega_c$$

It needs to happen that \$10\omega_c\leq SR\$ so that the output is not SR-limited.

In this case, for your 1MHz cutoff, \$10(2\pi f_c)\approx63V/\mu s\$. So your output will definitely SR-limited and this is just for a unit step input (your square wave has amplitude of 5V). In fact (theoretically) the output will not be SR-limited for values of the input of about 15mV or less. But you'd still have the BW limitation, which will keep the maximum slope at:

$$\dfrac{dv_o}{dt}\bigg|_{max}=10V_{in}\omega_c \text{  for sufficiently small }V_{in}$$

And when \$V_{in}\$ is big enough so that the previous equation is greater than the SR spec—then the limitation will be the SR. For practical purposes, you'd still be SR-limited because many opamps have offset voltages in the range of the minimum input voltage found in this problem (unless you use a precision opamp) but this is homework...