I don't understand your KVL equation. There's a term \$R_CI_E\$ which shouldn't appear in there, because it does not form part of the loop this equation is dealing with. Then, even though that term magically disappears in your rearrangement, everything's correct but you still get the wrong answer for \$I_B\$. You somehow got 6μA, when it should be 600μA.
Here's your circuit, annotated to help my explanation:
simulate this circuit – Schematic created using CircuitLab
Considering the loop consisting of nodes E, B and D, \$R_C\$ isn't part of it, and won't appear in any KVL equation we derive for that loop. There are a few ways to go about applying KVL here, but I'll start at node A, and go downwards and clockwise.
First we encounter a potential decrease \$V_{EB}\$ across the base emitter junction, which would make the KVL term negative. Then we have a further decrease of whatever voltage appears across \$R_B\$ (another negative term), and finally an increase in potential from -4.6V to 2V:
$$
\begin{aligned}
-V_{BE} - R_BI_B + \left[(+2)-(-4.6)\right] &= 0 \\ \\
-0.6 - R_BI_B + 6.6 &= 0 \\ \\
R_BI_B &= 6.6 - 0.6 \\ \\
&= 6.0 \\ \\
I_B &= \frac{6.0}{R_B} \\ \\
&= \frac{6.0}{10k} \\ \\
&= 600\mu A \\ \\
\end{aligned}
$$
All this assumes that the base-emitter junction is forward biased, which it is, and is not blocking current.
There are limits to the amount of collector current \$I_C\$ we can expect to see. If the transistor is off, obviously there's no collector current. It's not off, because \$I_B \ne 0\$. When the transistor is saturated, \$V_{CE} = 0.2V\$, and we can apply KVL again (to the loop consisting of nodes E, C and G) to find saturation current \$I_{CSAT}\$:
$$
\begin{aligned}
-0.2 - I_{CSAT}R_C + 2.0 &= 0 \\ \\
I_{CSAT} &= \frac{2.0 - 0.2}{R_C} \\ \\
&= \frac{1.8}{1k} \\ \\
&= 1.8mA
\end{aligned}
$$
Collector current cannot exceed 1.8mA. If, when we use the transistor's current gain to calculate \$I_C\$, we obtain more than 1.8mA, the transistor is saturated:
$$
\begin{aligned}
I_C &= \beta I_B \\ \\
&= 20 \times 600\mu A \\ \\
&= 12mA
\end{aligned}
$$
That's not possible, this transistor is saturated, and:
$$ I_C = 1.8mA $$