Proportional fair scheduling
I see how you came up with your result -- it is a different interpretation of the phrase "relatively better." It does not mean relatively better than the other user, but relatively better for the same user. 

To calculate the speeds, look at a given time slot and randomly select a speed for both user 1 and user 2. One of four cases should occur, with equal probability (since speed selection is 0.5 for each):


User 1 at 76.8 (low for U1), user 2 at 153.6 (low for U2). Since both are "relatively low", choose between the users with equal probability.
User 1 at 153.6 (high for U1), user 2 at 153.6 (low for U2). Always choose U1.
User 1 at 76.8 (low for U1), user 2 at 307.2 (high for U2). Always choose U2.
User 1 at 153.6 (high for U1), user 2 at 307.2 (high for U2). Tie, so choose between the two with equal probability.


The resulting average speed is then:

$$\text{User 1 speed} = \frac{1}{4}\left(\frac{1}{2}\cdot76.8+153.6+0+\frac{1}{2}\cdot153.6\right)\text{ kbps} = 67.2\text{ kbps}$$

$$\text{User 2 speed} = \frac{1}{4}\left(\frac{1}{2}\cdot153.6+0+307.2+\frac{1}{2}\cdot307.2\right)\text{ kbps} = 134.4\text{ kbps}$$